Sum up the degrees of the vertices in two ways, once as a sum over the vertices and once as a sum over the edges. Assume you don't have two vertices with degree one, use this assumption to get a lower bound for the sum of the degrees and you'll find a contradiction.

Well if you assumed you had no vertices of degree 1, then each [tex]d_i[/tex] must be greater than or equal to what number? What does this say about [tex]\sum d_i[/tex], keeping in mind that [tex]p[/tex] is the number of vertices?

Can we modify this slightly to rule out the possiblity of only one vertex of degree 1? What about 2 vertices of degree 1, is this possible?

Hang on, this lower bound of 2p came from the assumption that we had no vertices of degree 1, so this contradiction only tells you that you have at least one vertex of degree 1. The point of this cse was to give you a (slightly) simpler version to try first.

What inequality do you get if you assume you have only one vertex of degree 1? You should get a similar contradiction.

Now suppose you have 1 vertex with degree 1, the rest two or more. Without loss of generality assume [tex]d_p=1[/tex] and [tex]d_i\geq 2[/tex] for all 1<=i<p. Using this, what is a lower bound for the sum?

use induction on the number of edges. It is true if there is only one edge.

Now just collapse any edge, collapsing vertex a to vertex b. The result is true for the new graph by induction. If collapsing the edge changed the degree of b from 2 to one, it is obvious that either b or a also had degree one originally.

I.e. all edges, except for e, that used to end at either a or b now end at c, so we must add them. (Since G was a tree, only one edge, namely e, ended at both a and b, so a and b had no edges in common except for e.) Since the missing edge e counted as one for both deg(a) and deg(b), we must subtract it twice have our formula. i.e. deg(c) = deg(a)-1 + deg(b)-1.

Now deg(c) cannot be zero, since that would mean that G had only the one edge e.

Consequently, deg(c) ≥ 1, and if deg(c) =1, then either deg(a) or deg(b)=1.

All other vertices of G' have the same degree they had in G, since they fail to meet e.

Now by the inductive hypothesis, there are two vertices say x,y, of G' of degree one. If neither of these is c, we are done, since our collapsing process did not affect them, so the same two vertices on G also had degree one.

If it happens that say x = c, then deg(c) =1, so either a or b had degree 1 in G, say deg(a) = 1. Then in G we had at least two vertices of degree one, namely a and y.